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As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1.  In Fresnel’s biprism experiment is held in water instead of air, then what will be the effect on fringe width
•   Decreases
•   Increases
•   No effect
•   None of these
Solution
𝛽 = (𝑎 + 𝑏)𝜆/2𝑎(𝜇 − 1)𝛼 ,𝑖.𝑒.,𝛽 ∝ 𝜆/(𝜇 − 1)
When placed in water 𝛽′ ∝ 𝜆/𝜇′/ ( 𝜇/𝜇′−1)
𝑖.𝑒.,𝛽′ ∝ 𝜆 (𝜇−𝜇′) but < 𝜇
∴ 𝛽′/𝛽 = (𝜇 − 1) (𝜇 − 𝜇′)
∵ 𝜇′ > 1𝜆 ∴ 𝛽′ > 𝛽
𝑖.𝑒., the fringe width increases

Q2. The radiation pressure (in 𝑁/𝑚2) of the visible light is of the order of
•   10-2
•   10-4
•   10-6
•   10-8
Solution
In 1903, the American scientists Nicols and Hull measured the radiation pressure of visible light. It was found to be of the order of 7 × 10-6𝑁/𝑚2

Q3.   . Two stars are situated at a distance of 8 light year from the earth. These are to be just resolved by a telescope of diameter 0.25 m. If the wavelength of light used is 5000Å, then the distance between the stars must be
•   3 × 1010 m
•   3.35 × 1011 m
•   1.95 × 1011 m
•   4.32 × 1010 m
Solution
Limit of resolution of the telescope 𝑎 = 1.22𝜆 / 𝑎 = 𝑑/𝑥 Or 𝑑 = 1.22𝜆 𝑥 /𝑎 = (1.22 × 5 × 10-7 × 8 × 1016) /0.25 = 1.95 × 1011m

Q4.  In Young’s double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If 𝐼𝑚 be the maximum intensity, the resultant intensity I when they interfere at phase difference 𝜙 is given by
•   𝐼m/9 (4 + 5cos𝜙)
•   𝐼m /3 (1 + 2cos2 𝜙/2 )
•   𝐼m/5 (1 + 4cos2 𝜙/2 )
•   𝐼m/9 (1 + 8cos2 𝜙/2 )
Solution
Let 𝐴1 = 𝐴0, Then 𝐴2 = 2𝐴0
Intensity 𝐼 ∝ 𝐴2
Hence 𝐼1 = 𝐼0,𝐼2 = 4𝐼0
We have 𝐼 = 𝐼0 + 4𝐼0 + 2√𝐼0 × 4𝐼0 cos𝜙
For 𝐼max,cos𝜙 = 1
Hence 𝐼m = 9𝐼0 or 𝐼0 = 𝐼0/9
When phase difference is 𝜙 then
𝐼 = 𝐼0 + 4𝐼0 + 2√4𝐼02 cos𝜙
= 𝐼0 + 4𝐼0(1 + cos𝜙)
= 𝐼0 (1 + 8cos2 𝜙/2 ) [∵ 1 + cos𝜙 = 2cos2 𝜙/2 ]
= 𝐼m / 9 (1 + 8cos1 𝜙/2)

Q5. Air has refractive index 1.0003. The thickness of air column, which will have one more wavelength of yellow light (6000Å) than in the same thickness of vacuum is
•   2mm
•   2cm
•   2m
•   2km
Solution
2mm

Q6.  In Young’s double slit experiment, the 7th maximum wavelength 𝜆1 is at a distance 𝑑1 and that with wavelength 𝜆2 is at a distance 𝑑2.Then (𝑑1/𝑑2) is
•   (𝜆1/𝜆2)
•   (𝜆2/𝜆1)
• (𝜆12/𝜆22)
•   (𝜆22/𝜆12)
Solution
From 𝑥 = 𝑛𝜆 𝐷/𝑑
𝑑1 = 7𝜆1 𝐷/𝑑
𝑑2 = 7𝜆2 𝐷/𝑑
∴ 𝑑1/𝑑2 = 𝜆1/𝜆2

Q7. The angular resolution of a 10 cm diameter telescope at a wavelength of 5000Å is of the order of
Solution

Angular resolution = 1.22𝜆/𝑑
= (1.22 × 5000 × 10-10)/(10 × 10-2) = 6.1 × 10-6

Q8. What is the path difference of destructive interference
•   𝑛𝜆
• 𝑛(𝜆 + 1)
•   (𝑛 + 1)𝜆 /2
•   (2𝑛 + 1)𝜆 /2
Solution
For destructive interference path difference is odd multiple of 𝜆 2

Q9. A beam of electron is used in an 𝑌𝐷𝑆𝐸 experiment. The slit width is d. When the velocity of electron is increased, then

•   No interference is observed
•   Fringe width increases
•   Fringe width decreases
•   Fringe width remains same
Solution
Momentum of the electron will increase. So the wavelength (𝜆 = ℎ/𝑝) of electrons will decrease and fringe width decreases as 𝛽 ∝ 𝜆

Q10.  In Young’s double slit experiment, angular width of fringes is 0.20° for sodium light of wavelength 5890 Å. If complete system is dipped in water, then angular width of fringes becomes
•   0.11°
•   0.15°
•   0.22°
• 0.30°
Solution
Angular fringe width 𝜃 = 𝜆/𝑑 ⇒ 𝜃 ∝ 𝜆
𝜆𝑤 = 𝜆𝑎 /𝜇𝑤
So 𝜃𝑤 = 𝜃air/ 𝜇𝑤 = 0.20/ (4 /3) = 0.15° ## Want to know more

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: WAVE OPTICS QUIZ 17
WAVE OPTICS QUIZ 17